Question

800g ice of 0 degree Celsius is mixed with 200g of steam of 100 degree Celsius. What will be final temperature of mixture??​

Answer 1

Answer:

50 degree Celsius because if ice is mixed stream then their respective temperature will be half

Answer 2

The final temperature of mixture is $19.2^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$         ............(1)

where,

q = heat absorbed or released

= mass of ice = 800 g

$m_2$ = mass of steam = 200 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of ice= $0^oC$

$T_2$ = temperature of steam = $100^oC$

$c_1$ = specific heat of ice= $2.1J/g^0C$

$c_2$ = specific heat of steam= $1.996J/g^0C$

Now put all the given values in equation (1), we get

$800\times 2.1\times (T_{final}-0)=-[200\times 1.996\times (T_{final}-100)]$

$T_{final}=19.2^0C$

Thus final temperature of mixture will be $19.2^0C$